= Solution
Let $A=\pi r^2$ and choose the rotating area normal to coincide initially with $+z$. At time $t$ its angle to the applied <magnetic field> is $\omega t$, so the external <magnetic flux> is $\Phi_B=BA\cos\omega t$. <Faraday's law> therefore gives
$$
\boxed{\mathcal E(t)=-\frac{d\Phi_B}{dt}=B\pi r^2\omega\sin\omega t}.
$$
In the high-<electrical resistance> approximation, neglecting the loop's <self-inductance>, <Ohm's law> gives
$$
\boxed{I(t)=\frac{B\pi r^2\omega}{R}\sin\omega t}.
$$
The displayed signs specify positive circulation by the chosen normal; reversing the reference circulation reverses both signs. The high-<electrical resistance> assumption is quantitatively $\omega L\ll R$ if the loop has <self-inductance> $L$. Otherwise the current lags the induced <electromotive force> and must be obtained from $L\dot I+RI=\mathcal E$.
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